Find the sum of the coefficients of ##(x+y)^{16}##

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  • #1
RChristenk
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Homework Statement
Find the sum of the coefficients of ##(x+y)^{16}##
Relevant Equations
Binomial Theorem
##(x+y)^{16}=[x(1+\dfrac{y}{x})]^{16}=x^{16}(1+\dfrac{y}{x})^{16}##

## x^{16}(1+\dfrac{y}{x})^{16}=x^{16}[1+^{16}C_1(\dfrac{y}{x})+^{16}C_2(\dfrac{y}{x})^2...+^{16}C_{16}(\dfrac{y}{x})^{16}]##

Now let ##x=1,y=1##:

##1^{16}(1+1)^{16}=1^{16}(1+^{16}C_1+^{16}C_2...+^{16}C_{16})##

##2^{16}-1=^{16}C_1+^{16}C_2...+^{16}C_{16}##

Sum of coefficients = ##2^{16}-1## = ##65535##

But the answer is ##65536##. Why?
 
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  • #2
RChristenk said:
Sum of coefficients = ##2^{16}-1## = ##65535##
The first '=' in that line is not correct. It does not follow from anything written above it.
You don't need most of the working in the OP. The sum of the coefficients will simply be the value of the expression when ##x=y=1##, since all items ##x^k y^{16-k}## will be 1. Hence the sum of the coefficients will just be ##(1+1)^{16}##.
 
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  • #3
andrewkirk said:
The first '=' in that line is not correct. It does not follow from anything written above it.
You don't need most of the working in the OP. The sum of the coefficients will simply be the value of the expression when ##x=y=1##, since all items ##x^k y^{16-k}## will be 1. Hence the sum of the coefficients will just be ##(1+1)^{16}##.
Could you tell me where I went wrong specifically? Because to me what I wrote down looks correct (although obviously it isn't). I know I could just set everything to 1 and plug it in, but then I don't really understand what's happening and it becomes a rote memory item to me.
 
  • #4
RChristenk said:
Could you tell me where I went wrong specifically? Because to me what I wrote down looks correct (although obviously it isn't). I know I could just set everything to 1 and plug it in, but then I don't really understand what's happening and it becomes a rote memory item to me.
You moved the first coefficient to the left hand side and gave the answer as ##N -1## rather than ##N##. Only you can explain why you did this!
 
  • #5
To rephrase @PeroK #4 :

RChristenk said:
Could you tell me where I went wrong specifically?

You overlooked that this number one is also a coefficient (##^{16}C_0## )

(easy check: same exercise with powers 0, 1, 2, ... instead of 16 :smile:)

##\ ##
 
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  • #6
PeroK said:
You moved the first coefficient to the left hand side and gave the answer as ##N -1## rather than ##N##. Only you can explain why you did this!
Uh..I thought ##C## represented the "C"oefficients, so naturally the digit one is moved to the other side. Now I see that is incorrect.
 
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1. How do you find the sum of the coefficients of (x+y)^16?

To find the sum of the coefficients of (x+y)^16, you simply need to substitute x=1 and y=1 into the expression and evaluate it. This is because the sum of the coefficients is equal to the value of the expression when all variables are set to 1.

2. What is the general formula for finding the sum of the coefficients of (x+y)^n?

The general formula for finding the sum of the coefficients of (x+y)^n is 2^n. This is because each term in the expansion represents a combination of choosing either x or y from each of the n factors, resulting in 2^n possible combinations, each with a coefficient of 1.

3. Why is it important to find the sum of the coefficients of a binomial expansion?

Finding the sum of the coefficients of a binomial expansion can provide valuable information about the properties of the expression, such as the total number of terms or the total value of the expression under certain conditions. It can also help simplify the expression or solve specific problems related to the expansion.

4. Can the sum of the coefficients of (x+y)^16 be calculated without expanding the expression?

Yes, the sum of the coefficients of (x+y)^16 can be calculated without expanding the expression by using the general formula mentioned earlier. By substituting n=16 into the formula 2^n, you can directly find the sum of the coefficients as 2^16=65536.

5. How does the sum of the coefficients of (x+y)^16 relate to Pascal's Triangle?

The sum of the coefficients of (x+y)^16 is equal to the sum of the numbers in the 16th row of Pascal's Triangle. This is because each row of Pascal's Triangle corresponds to the coefficients of the binomial expansion of (x+y)^n, where n is the row number. Therefore, the sum of the coefficients of (x+y)^16 is equal to the sum of the numbers in the 16th row of Pascal's Triangle, which is 2^16=65536.

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